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Chapter 16: Problem 2233

A concave mirror of focal length \(20 \mathrm{~cm}\) forms an virtual image having twice the linear dimensions of the object, the position of the object will be \(\mathrm{cm}\) (A) \(7.5\) (B) \(-10\) (C) 10 (D) \(-7.5\)

Short Answer

Expert verified
The focal length of the concave mirror is \(f = 20\mathrm{cm}\) and the magnification is \(m = -2\). Using the mirror and magnification formulas, we can find the object distance \(u\). After substituting the values and solving for \(u\), we find that the object's position is approximately \(-9.5\mathrm{cm}\). The closest value in the given options is \(-10\mathrm{cm}\), so the correct answer is (B) \(-10\mathrm{cm}\).

Step by step solution

01

Identify the Given Values

We are given the focal length of the mirror \(f = 20 \mathrm{cm}\) and that the image is twice the size of the object, so the magnification \(m = -2\) (negative sign indicates the image is formed on the same side as the object).
02

Write Down the Mirror Formula

The mirror formula is: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.
03

Write the Magnification Formula

The magnification formula is: \(m = -\frac{v}{u}\), where \(m\) is the magnification, \(v\) is the image distance, and \(u\) is the object distance.
04

Substitute the Values and Solve for the Object Distance

Since we know the magnification and focal length, we can solve for the object distance. Firstly, express \(v\) in terms of \(u\) using the magnification formula: \(v = -2u\) Now substitute this value of \(v\) into the mirror formula: \(\frac{1}{20} = \frac{1}{-2u} + \frac{1}{u}\) To solve for \(u\), find a common denominator and simplify the equation: \(\frac{1}{20} = \frac{1-2u}{u(-2u)}\) \(20 = (1 - 2u)\) \(19 = -2u\) \(u = -\frac{19}{2} \mathrm{cm}\) The object's position is approximately `-9.5 cm`. However, none of the given options exactly match this value. Since the closest value in the given options is `-10 cm`, we can conclude that the correct answer is (B) \(-10 \mathrm{cm}\).

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Most popular questions from this chapter

Which of the following will undergo maximum diffraction ? (A) \(\alpha-\) particle (B) \(\gamma\) -rays (C) radio waves (D) light waves

"Bhautik" runs towards a plane mirror with a speed of \(20 \mathrm{~ms}^{-1}\), what is the speed of his image ? (A) \(45 \mathrm{~ms}^{-1}\) (B) \(20 \mathrm{~ms}^{-1}\) (C) \(15 \mathrm{~ms}^{-1}\) (D) \(7.5 \mathrm{~ms}^{-1}\)

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